Question 1209372
<pre>
Trial and error may not be obvious to you, so you
might try doing a little algebra to make it easier
to see what to do.  

{{{matrix(2,3,

3^x+3^y,""="",108,
3^x+3^y,""="",2^2*3^3)}}}

Divide both sides by 3<sup>3</sup> by subtracting
3 from both exponents of 3 on the left:

{{{matrix(2,3,
3^(x-3)+3^(y-3),""="",2^2,
3^(x-3)+3^(y-3),""="",4)}}}

Factor out 3<sup>x-3</sup> on the left:

{{{matrix(2,3,
3^(x-3)(1+3^(y-3-x+3)),""="",4,
3^(x-3)(1+3^(y-x)),""="",4)}}}

That reminds us a little of {{{1*(1+3)}}}{{{""=""}}}{{{4}}}.

That would be the case if the power of 3 before the parentheses 
were equal to 1, and the power of 3 inside the parentheses were
3<sup>1</sup>.  

Now we know that 3<sup>0</sup> = 1.  And we see that x = 3 would make that
exponent = 0.

So now all we need to do is find y so that y-x = 1.  Substituting 
x = 3,  y - 3 = 1, or y = 4.

So  x = 3 and y = 4 are the answers.  Checking:

{{{matrix(4,3,

3^x+3^y,""="",108, 
3^3+3^4,""="",108,
27+81,""="",108,
108,""="",108)}}}

Edwin</pre>